A solution of potassium hydroxide is in equilibrium with an undissolved solute at 45 degrees Celsius. What will happen to the mass of the potassium hydroxide if the temperature is raised to 50 degrees Celsius? (Given that heat of solution is -57.6kJ/mol)Please explain as well.
To determine what will happen to the mass of potassium hydroxide (KOH) when the temperature is raised from 45 degrees Celsius to 50 degrees Celsius, we need to consider the effect of temperature on the solubility of KOH.
The solubility of most solid solutes in water generally increases with increasing temperature. In this case, since we have a solute (undissolved KOH) in equilibrium with a solution of KOH, increasing the temperature will cause more KOH to dissolve.
Now, to understand why this happens, we can look at the concept of Le Chatelier's principle. According to this principle, when a system at equilibrium is subjected to a change (such as a change in temperature), the system will adjust to minimize the effect of that change. In this case, increasing the temperature would be considered a stress on the equilibrium.
Since the dissolution of KOH is an exothermic process (heat is released), increasing the temperature would be equivalent to adding heat to the system. According to Le Chatelier's principle, the system will respond to this increase in temperature by shifting in the direction that absorbs heat, which in this case is the dissolution of more KOH.
Therefore, when the temperature is raised from 45 degrees Celsius to 50 degrees Celsius, more KOH will dissolve, resulting in an increase in the mass of the potassium hydroxide.
To calculate the change in mass, we would need additional information such as the initial mass of undissolved KOH and the concentration of the solution.